How do you solve the equation $6 x^{2} - 5 x - 13 = x^{2} - 11$ $6x^2-5x-13=x^2-11$ by completing the square?

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How do you solve the equation $6 x^{2} - 5 x - 13 = x^{2} - 11$ $6x^2-5x-13=x^2-11$ by completing the square?

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Answer 1 · 2017-08-20T02:50:33

$x = \sqrt{\frac{13}{20}} + \frac{1}{2}$ $x=sqrt(13/20)+1/2$
$x = - \sqrt{\frac{13}{20}} + \frac{1}{2}$ $x=-sqrt(13/20)+1/2$

Explanation:

Given -

$6 x^{2} - 5 x - 13 = x^{2} - 11$ $6x^2-5x-13=x^2-11$

Take all the terms to the left-hand side

$6 x^{2} - 5 x - 13 - x^{2} + 11 = 0$ $6x^2-5x-13-x^2+11=0$

Simplify it.

$5 x^{2} - 5 x - 2 = 0$ $5x^2-5x-2=0$

Take the constant term to the right-hand side

$5 x^{2} - 5 x = 2$ $5x^2-5x=2$

Divide all the terms by the coefficient of $x^{2}$ $x^2$

$\frac{5 x^{2}}{5} - \frac{5 x}{5} = \frac{2}{5}$ $(5x^2)/5-(5x)/5=2/5$

$x^{2} - x = \frac{2}{5}$ $x^2-x=2/5$

Take half the coefficient of $x$ $x$ , square it and add it to both sides

$x^{2} - x + \frac{1}{4} = \frac{2}{5} + \frac{1}{4} = \frac{8 + 5}{20} = \frac{13}{20}$ $x^2-x+1/4=2/5+1/4=(8+5)/20=13/20$

${(x - \frac{1}{2})}^{2} = \frac{13}{20}$ $(x-1/2)^2=13/20$

$(x - \frac{1}{2}) = \pm \sqrt{\frac{13}{20}}$ $(x-1/2)=+-sqrt(13/20)$

$x = \sqrt{\frac{13}{20}} + \frac{1}{2}$ $x=sqrt(13/20)+1/2$
$x = - \sqrt{\frac{13}{20}} + \frac{1}{2}$ $x=-sqrt(13/20)+1/2$

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How do you solve the equation $6 x^{2} - 5 x - 13 = x^{2} - 11$ $6x^2-5x-13=x^2-11$ by completing the square?

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How do you solve the equation $6 x^{2} - 5 x - 13 = x^{2} - 11$ $6x^2-5x-13=x^2-11$ by completing the square?